What is the integral of ((x^2))/(((x-1)(x^2+2x+1))) ?

The integral of ((x^2))/(((x-1)(x^2+2x+1))) is 1/4 ln|x-1|+3/4 ln|x+1|+1/(2(x+1))+C

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integral of ((x^2))/(((x-1)(x^2+2x+1)))

\int \frac{( x ^{2} )}{( ( x - 1 ) ( x ^{2} + 2 x + 1 ) )} d x

\frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + C

Solution steps

\int \frac{x ^{2}}{( x - 1 ) ( x ^{2} + 2 x + 1 )} d x

Take the partial fraction of

\frac{x ^{2}}{( x - 1 ) ( x ^{2} + 2 x + 1 )} : \frac{1}{4 ( x - 1 )} + \frac{3}{4 ( x + 1 )} - \frac{1}{2 ( x + 1 ) ^{2}}

= \int \frac{1}{4 ( x - 1 )} + \frac{3}{4 ( x + 1 )} - \frac{1}{2 ( x + 1 ) ^{2}} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{4 ( x - 1 )} d x + \int \frac{3}{4 ( x + 1 )} d x - \int \frac{1}{2 ( x + 1 ) ^{2}} d x

\int \frac{1}{4 ( x - 1 )} d x = \frac{1}{4} ln ∣ x - 1 ∣

\int \frac{3}{4 ( x + 1 )} d x = \frac{3}{4} ln ∣ x + 1 ∣

\int \frac{1}{2 ( x + 1 ) ^{2}} d x = - \frac{1}{2 ( x + 1 )}

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ - (- \frac{1}{2 ( x + 1 )})

Simplify

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )}

Add a constant to the solution

= \frac{1}{4} ln ∣ x - 1 ∣ + \frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + C

\frac{d}{d x} (\frac{1}{cos ( x )})

\frac{s + 1}{( s + 2 ) ( s + 3 )}

x \to 0 + lim (ln (\frac{7}{x ^{2}}))

n = 1 \sum \infty (\frac{3}{8})^{1 - n}

x \to 1 lim (e^{6 x^{4} - 6 x})

What is the integral of ((x^2))/(((x-1)(x^2+2x+1))) ?
The integral of ((x^2))/(((x-1)(x^2+2x+1))) is 1/4 ln|x-1|+3/4 ln|x+1|+1/(2(x+1))+C